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marissa [1.9K]
3 years ago
12

Two ligands, a and b, both form complexes with a particular metal ion. when the metal ion complexes with ligand a, the resulting

solution is red. when the metal ion complexes with ligand b, the resulting solution is yellow. which of the two ligands produces the larger δ? two ligands, and , both form complexes with a particular metal ion. when the metal ion complexes with ligand , the resulting solution is red. when the metal ion complexes with ligand , the resulting solution is yellow. which of the two ligands produces the larger ? ligand a produces a higher δ ligand b produces a higher δ ligands a and b produce the same δ there is not enough data to determine.
Chemistry
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

because

Explanation:

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A 10.0 gram sample of Fe contains how many miles of Fe
Dafna1 [17]

Answer:

1.7857 moles

Explanation:

moles=mass/Mr

10/56=1.7857

moles of iron =1.7857

4 0
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How many moles of hcl are required to neutralize aqueous solutions of these bases? 0.03 mol koh?
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 The  moles  of  HCl    are  required  to  neutralized    aqueous     solution     of  the   0.03  KoH

KOH  +  HCl  =  KCl  +   H2O


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5 0
3 years ago
According to the aufbau principle,_____.
Softa [21]

Answer:

Hi there!

The correct answer to this question is: electrons enter orbitals of lowest energy first.

5 0
3 years ago
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Planet P has a rocky surface and a radius of 9250 km. At height h above the surface, the gravitational acceleration is What is h
cricket20 [7]
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5 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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